Question: Circle $\omega$ has radius 5 and is centered at $O$. Point $A$ lies outside $\omega$ such that $OA=13$. The two tangents to $\omega$ passing through $A$ are drawn, and points $B$ and $C$ are chosen on them (one on each tangent), such that line $BC$ is tangent to $\omega$ and $\omega$ lies outside triangle $ABC$. Compute $AB+AC$ given that $BC=7$.

[asy]

unitsize(0.1 inch);

draw(circle((0,0),5));
dot((-13,0));
label("$A$",(-13,0),S);

draw((-14,-0.4)--(0,5.5));
draw((-14,0.4)--(0,-5.5));

draw((-3.3,5.5)--(-7.3,-5.5));

dot((0,0));
label("$O$",(0,0),SE);

dot((-4.8,1.5));
label("$T_3$",(-4.8,1.5),E);

dot((-1.7,4.7));
label("$T_1$",(-1.7,4.7),SE);

dot((-1.7,-4.7));
label("$T_2$",(-1.7,-4.7),SW);

dot((-3.9,3.9));
label("$B$",(-3.9,3.9),NW);

dot((-6.3,-2.8));
label("$C$",(-6.3,-2.8),SW);

[/asy]
Solution: Let $T_1, T_2$, and $T_3$ denote the points of tangency of $AB, AC,$ and $BC$ with $\omega$, respectively.


[asy]

unitsize(0.1 inch);

draw(circle((0,0),5));
dot((-13,0));
label("$A$",(-13,0),S);

draw((-14,-0.4)--(0,5.5));
draw((-14,0.4)--(0,-5.5));

draw((-3.3,5.5)--(-7.3,-5.5));

dot((0,0));
label("$O$",(0,0),SE);

dot((-4.8,1.5));
label("$T_3$",(-4.8,1.5),E);

dot((-1.7,4.7));
label("$T_1$",(-1.7,4.7),SE);

dot((-1.7,-4.7));
label("$T_2$",(-1.7,-4.7),SW);

dot((-3.9,3.9));
label("$B$",(-3.9,3.9),NW);

dot((-6.3,-2.8));
label("$C$",(-6.3,-2.8),SW);

[/asy]

Then $7 = BC=BT_3+T_3C = BT_1 + CT_2$. By Pythagoras, $AT_1 = AT_2 = \sqrt{13^2-5^2}=12$. Now note that $24 = AT_1 + AT_2 = AB + BT_1 + AC + CT_2 = AB+AC+7$, which gives $AB + AC = \boxed{17}$.